package cn.lbd.arithmetic.leetcode.editor.cn;
//给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数 。
//
// 你可以对一个单词进行如下三种操作： 
//
// 
// 插入一个字符 
// 删除一个字符 
// 替换一个字符 
// 
//
// 
//
// 示例 1： 
//
// 输入：word1 = "horse", word2 = "ros"
//输出：3
//解释：
//horse -> rorse (将 'h' 替换为 'r')
//rorse -> rose (删除 'r')
//rose -> ros (删除 'e')
// 
//
// 示例 2： 
//
// 输入：word1 = "intention", word2 = "execution"
//输出：5
//解释：
//intention -> inention (删除 't')
//inention -> enention (将 'i' 替换为 'e')
//enention -> exention (将 'n' 替换为 'x')
//exention -> exection (将 'n' 替换为 'c')
//exection -> execution (插入 'u')
// 
// Related Topics 字符串 动态规划 
// 👍 1222 👎 0


//leetcode submit region begin(Prohibit modification and deletion)
class Solution72 {

    //自顶向下的递归，重复计算子问题，超时
    /*public static int minDistance(String word1, String word2) {
        //递归出口
        if (word1.length() == 0 || word2.length() == 0) {
            return Math.max(word1.length(), word2.length());
        }

        //最后字符相同
        if (word1.charAt(word1.length() - 1) == word2.charAt(word2.length() - 1)) {
            return minDistance(word1.substring(0, word1.length() - 1), word2.substring(0, word2.length() - 1));
        }

        //增删改取最小
        return Math.min(Math.min(minDistance(word1, word2.substring(0, word2.length() - 1)), minDistance(word1.substring(0, word1.length() - 1), word2))
                , minDistance(word1.substring(0, word1.length() - 1), word2.substring(0, word2.length() - 1))) + 1;
    }*/

    //自底向上的动态规划
    public static int minDistance(String word1, String word2) {
        int[][] dp = new int[word1.length() + 1][word2.length() + 1];
        //B串为空，A要删除i次
        for (int i = 1; i <= word1.length(); i++) {
            dp[i][0] = i;
        }
        //A串为空，A要添加j次
        for (int j = 1; j <= word2.length(); j++) {
            dp[0][j] = j;
        }

        for (int i = 1; i <= word1.length(); i++) {
            for (int j = 1; j <= word2.length(); j++) {
                //字符相等，不用做改变，上一个状态和当前状态保持一致
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    //从增删改的记忆集里求一个最小值
                    dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
                }
            }
        }
        return dp[word1.length()][word2.length()];
    }

    public static void main(String[] args) {
        String word1 = "horse";
        String word2 = "ros";
        System.out.println(minDistance(word1, word2));
    }
}
//leetcode submit region end(Prohibit modification and deletion)
